Q2 Flashcards

1
Q

Needs/purpose for electrical isolation

A
  • Isolate low dc voltages from high grid voltages (improving safety for user)
  • Transformer allow us to convert voltage levels between primary and secondary to reduce I and V stress of power devices (when large dc-dc conversion needed -i.e. using flyback instead of boost)
  • To break loops and separate parts of circuit sensitive to noise.
  • To have different reference potentials with respect to the input, output V is fixed but output can be shifted or inverted.
  • transformer of isolated converters operate at high frequencies, which reduces its size and weight->(high power density)
  • More efficient than linear power supplies, dc-dc isolated converters operate the devices in the ohmic region not in active region.
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2
Q

Problems caused by leakage inductance

A
  • Energy stored by leakage L must be absorbed by power devices, increasing stress on the devices and power losses
  • Additional components needed to dissipate leakage energy and protect power devices
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3
Q

Disadvantage of HF transformers

A
  • Not perfect coupling between windings can be achieved, resulting in a leakage L.
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4
Q

Alternatives to reduce leakage L

A
  • Good transformer design
  • Passive snubbers (RCD)
  • Active snubbers
  • Using leakage L for the operation of the converter.
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5
Q

Explain general ideas and limitations of good design of transformer

A
  • Maximize coupling between primary and secondary
  • Minimizing distance between primary and secondary windings
  • i.e. in flyback transformer we can use this structure where primary winding is split into 2 layers to increase coupling
  • Leakage L cannot be removed completely
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6
Q

Explain general ideas of passive snubers (RCD in flyback)

A
  • Snubbers are used to absorb/dissipate leakage energy in a resistive element
  • RCD snubber in flybacks limits voltage stress in the MOSFET and dissipates leakage energy in the R
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7
Q

explain general idea of active snubber in flyback

A
  • Active snubber allow recycling the leakage energy and reduce power losses
  • i.e. in flyback when MOSFET Q1 is turned off, leakage I flows through body diode of Q2 storing energy in clamp C, then clamp C starts to resonate with leakage L, and MOSFET Q2 is turned on so that leakage I changes direction and recycles energy to the output.
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8
Q

Explain idea of using leakage L in resonant converters

A
  • Leakage L can be used for the operation of the converter, known as resonant converter
  • By using resonant converter it’s possible to achieve zero-voltage or zero-current switching reducing switching losses and losses related to leakage L.
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9
Q

Q3. flyback CCM main ideas

A
  • In CCM characteristics
  • Minimum primary L
  • Voltage across primary side inductor ON/OFF state, determines rate of change of magnetizing I
  • Voltage across MOSFET/diode ON/OFF state
  • Current of MOSFET/diode ON/OFF state
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10
Q

Characteristics and condition for CCM operation of flyback

A
  • Only a part of energy stored in flyback transformer delivered to output, some energy remains when next cycle starts
  • To operate in CCM, Lp should be large enough to ensure magnetizing I never reach zero in the OFF interval
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11
Q

Definition of DCM operation of flyback

A
  • All energy stored in transformer is transferred to the load in the off period
  • Magnetizing I goes to zero
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12
Q

Advantages of DCM

A
  • Lp value is smaller reducing size of transformer
  • Soft switching of secondary diode, I decreases with slope -Vo/Ls and not abruptly, it might lead to less recovered charge and losses.
  • Turn-on of MOSFET is softer since I increases from zero with slope Vi/Lp and not abruptly, it might lead to less switching losses.
  • Possibility of implement QR resonant flyback due to resonance between Lp and output C of MOSFET
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13
Q

Disadvantages of DCM

A
  • Larger ripple and peak currents
  • Higher MOSFET turn off losses due to higher peak I
  • Higher leakage energy that has to be dissipated due to higher peak I
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14
Q

Advantages of CCM

A
  • Smaller ripple I and peak I
  • Lower MOSFET turn-off losses
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15
Q

Disadvantages of CCM

A
  • Inductance value for CCM is higher, leading to bigger transformer
  • Hard-switching of diode, leading to increased reverse-recovery losses due to high di/dt and reverse peak I.
  • Needs slope compensation for duty cycles higher than 50%
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16
Q

Steps to find voltage gain calculation in CCM

A
  • Find voltage across Lp when switch is ON
  • Find voltage across Lp when swtich is OFF
  • Apply second-volt balance of Lp, adding areas under curve of Vl and equating them to zero.
17
Q

Steps to find voltage gain in DCM and observations

A
  • Consider all energy stoed in Lp is transferred to output during each switching period
  • Calculate the average input power dividing energy stored in Lp by Ts
  • Find primary peak I multiplying slope Vin/Lp by the time the switch is ON (DTs)
  • Voltage gain expression is more complex and depends on fs, output load, Lp.
18
Q

Advantages of flyback for multiple outputs and limitations

A
  • Flyback can have multiple outputs from a single input source.
  • Output voltages will be proportional to the turns ratio of each winding.
  • Ideally we only need to regulate 1 output and the other will scale by the number of turns.
  • Parasitic elements affect load regulation of unregulated outputs.
19
Q

Explain how cross regulation is affected in a multiple output flyback

A
  • Occurs when 1 or more outputs are at minimum or zero load, and 1 output is at maximum load.
  • When MOSFET is turned off, leakage L in the primary induces a voltage spike common to all windings.
  • With no output load the C is charged to the peak voltage, until the diode stops conducting and C cannot discharge.
  • If load of 1 output keeps increasing, voltage spike caused by leakage L increases and voltage of output with no load increase.
20
Q

Control techniques for multiple output flyback and general explanation

A
  • AC stacked windings and combined feedback
  • Output 1 voltage will be proportional to (m+n) and output 2 voltage will be proportional to m
  • combine the outputs connecting them with a R to reference pin of shunt regulator
  • Regulator controls combination of outputs with only one feedback point, so no output will be perfectly regulated
  • i.e. if W1=0.9 and W2=0.1, output 1 will be 9 times more important than output 2, and will have better regulation.
21
Q

Explain procedure to calculate R of combined feedback method

A
  • We assign an importance/weight to each output W1 and W2
  • total I through R0 will be contribution of each output multiplied by the weight
  • we know that i1 will be difference between output and Vref times R1, and solve for R1
  • We do the same for R2